Question

Giải phương trình

\(2\left|sinx+cosx\right|+3sin2x=2\)

Answer 1

Đặt \(\left|sinx+cosx\right|=a\) (\(0\le a\le\sqrt{2}\))

\(\Rightarrow1+2sinx.cosx=a^2\Leftrightarrow1+sin2x=a^2\Rightarrow sin2x=a^2-1\)

Phương trình trở thành:

\(2a+3\left(a^2-1\right)=2\Leftrightarrow3a^2+2a-5=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{5}{3}< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left|sinx+cosx\right|=1\Leftrightarrow\left|sin\left(x+\frac{\pi}{4}\right)\right|=\frac{\sqrt{2}}{2}\)

\(\Rightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\\sin\left(x+\frac{\pi}{4}\right)=\frac{-\sqrt{2}}{2}\end{matrix}\right.\) \(\Rightarrow...\)